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3 years ago

What number is 24% of 80

Mathematics

2 answers:

Anna [14]3 years ago

4 0

Percent means parts out of 100
24%=24/100=2.4/10=0.24/1=0.24
'of' means multiply
24% of 80=0.24 times 80=19.2

ch4aika [34]3 years ago

3 0

19.2 is 24% of 80 when i did the work

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Will someone help me with both of these I have to solve for the unknown portions please

IceJOKER [234]

Answer:

b=2.5

p=6

Step-by-step explanation:

Take the fraction and simplify it. So 10/8 would be 5/4

You multiply the 2 by 2, giving you h/4

So now you have 5/4=h/4

h=5 but now you have to divide both by 2 again so it can go back to the original fraction.

5÷2=2.5 and 4÷2=2

Second Problem! 4/2 is equal to 2/1

You can also multiply 4 by 3 to get the numerator as 12. Multiply the 4 by 3 and the 2 by 3

Now you have 12/6= 12/p

So p is 6

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Soloha48 [4]

The volume of a box is equal to the product its width, length and height
V = whl
Since all of the sides are 4 inches, we end up with
V= 4*4*4
Simplify to V = 16*4 = 64.
If the volume of one box is 64:
V = 64
The volume of 12 boxes is 64 * 12.
12V = 768
Since the volume of 12 boxes is the same as the volume of the carton, the volume of the carton must be 768 cubic inches.

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3 years ago

A counting number is a multiple of 13 and is also divisible by 4 and 6. What is the smallest number that satisfies these conditi

ASHA 777 [7]

Answer:

Step-by-step explanation:

39 is a multiple of 13. 13,26,39.

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In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

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3 years ago